If the percentage of Se (atomic weight 78.4) | vedclass

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(A) The minimum molecular weight is calculated assuming at least one atom of $Se$ is present in the enzyme molecule.Given: Percentage of $Se = 0.5\%$,

Vedclass · Accepted Answer

$1.568 	imes 10^4$

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(A) The minimum molecular weight is calculated assuming at least one atom of $Se$ is present in the enzyme molecule.Given: Percentage of $Se = 0.5\%$,Atomic weight of $Se = 78.4$.Formula: $	ext{Minimum Molecular Weight} = \frac{	ext{Atomic weight of element} 	imes 100}{	ext{Percentage of element}}$.Calculation: $	ext{Molecular Weight} = \frac{78.4 	imes 100}{0.5} = \frac{7840}{0.5} = 15680$.In scientific notation,this is $1.568 	imes 10^4$.

If the percentage of $Se$ (atomic weight $78.4$) in a peroxidase anhydrous enzyme is $0.5\%$,what is the minimum molecular weight of the peroxidase anhydrous enzyme?

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